with these directions is altered by the strain into a rhombus whose diagonals are (1+e) 2 and (1-e)/2, and whose area, being half the product of the diagonals, is 12, or, to the first order of small quantities, is 1, the same as the area of the original square. The length of a side of the rhombus, being the square root of the

sum of the squares of the semi-diagonals, is found to be √ 1 + e2 or 1 + 12, and is therefore, to the first order of small quantities, equal to a side of the original square.

I

56. To find the magnitude of the small angle which a side of the rhombus makes with the corresponding side of the square, we may proceed as follows:-Let acb (Fig. 2) be an enlarged representation of one of the small triangles in Fig. 1. Then we have ab = }, cb = }e √2 =

e

√2√2

equal to ab, we have, to the first order of small quan

The semi-angles of the rhombus are therefore± e,

π

4

and the angles of the rhombus are 2e; in other

2

words, each angle of the square has been altered by the amount 2e. This quantity 2e is adopted as the measure of the shear.

57. To find the perpendicular distance between opposite sides of the rhombus, we have to multiply a side by the cosine of 2e, which, to the first order of small quantities, is 1. Hence the perpendicular distance between opposite sides of the square is not altered by the shear, and the relative movement of these sides is represented

by supposing one of them to remain fixed, while the other slides in the direction of its own length through a distance of ze, as shown in Fig. 3 or Fig. 4. Fig. 3, in fact, represents a shear combined with right-handed rotation, and Fig. 4 a shear combined with left-handed rotation, as appears by comparing these figures with Fig. 1, which represents shear without rotation.

58. The square and rhombus in these three figures may be regarded as sections of a prism whose edges are perpendicular to the plane of the paper, and figures 3 and 4 show that (neglecting rotation) a shear consists in the

D

relative sliding of parallel planes without change of distance, the amount of this sliding being proportional to the distance, and being in fact equal to the product of the distance by the numerical measure of the shear. A good illustration of a shear is obtained by taking a book, and making its leaves slide one upon another.

It may be well to remark, by way of caution, that the selection of the planes is not arbitrary as far as direction is concerned. The only planes which are affected in the manner here described are the two sets of planes which make angles of 45° with the axes of the shear (these axes being identical with the diagonals in Fig. 1).

59. Having thus defined and explained the term "shear," which it will be observed denotes a particular species of strain, we now proceed to define a shearing

stress.

A shearing stress may be defined as the combination of two longitudinal stresses at right angles to each other, these stresses being opposite in sign and equal in magnitude; in other words, it consists of a pull in one direction combined with an equal thrust in a perpendicular direction.

D

C

60. Let P denote the intensity of each of these longitudinal stresses; we shall proceed to calculate the stress upon a plane inclined at 45° to the planes of these stresses. Consider a unit cube so taken that the pull is perpendicular to two of its faces, A B and DC (Fig. 5), and the thrust is perpendicular to two other faces, AD, BC. The forces which hold the half-cube ABC in equilibrium are—

Fig 5.

(1) An outward force P, uniformly distributed over the face A B, and having for its resultant a single force P acting outward applied at the middle point of AB.

(2) An inward force P, having for its resultant a single force P acting inwards at the middle point of B C.

(3) A force applied to the face A C.

To determine this third force, observe that the other two forces meet in a point, namely the middle point of AC, that their components perpendicular to A C destroy one another, and that their components along A C, or

Р

rather along C A, have each the magnitude ; hence

√2 their resultant is a force P√2, tending from C towards A. The force (3) must be equal and opposite to this. Hence each of the two half-cubes A B C, A D C exerts upon the other a force P/2, which is tangential to their plane of separation. The stress upon the diagonal plane A C is therefore a purely tangential stress. To compute its intensity, we must divide its amount P√2 by the area of the plane, which is √2, and we obtain the quotient P. Similar reasoning applies to the other diagonal plane B D. P is taken as the measure of the shearing stress. The above discussion shows that it may be defined as the intensity of the stress either on the planes of purely normal stress, or on the planes of purely tangential stress.

61. A shearing stress, if applied to a body which has the same properties in all directions (an isotropic body), produces a simple shear with the same axes as the stress; for the extension in the direction of the pull will be equal to the compression in the direction of the thrust; and in the third direction, which is perpendicular to both of these, there is neither extension nor contraction, since

the transverse contraction due to the pull is equal to the transverse extension due to the thrust.

A shearing stress applied to a body which has not the same properties in all directions, produces in general a shear with the same axes as the stress, combined with some other distortion.

In both cases, the quotient of the shearing stress by the shear produced is called the resistance to shearing. In the case of an isotropic body, it is also called the simple rigidity.

62. The following values of the resilience of liquids under compression are reduced from those given in Jamin, Cours de Physique,' 2nd edition, tom. i. pp. 168 and 169:

: